Newton’s 3rd Law of Motion
Calculation of Momentum
Momentum is a new quantity that is simply mass x velocity (p = mv). It is a vector
quantity and has the standard units of kgm/s. Momentum can be thought of a
moving inertia. Why does an object in motion stay in motion? Because it has
momentum. It takes an unbalanced force to increase or decrease an object’s
momentum. The greater the mass the greater the force required in to change that
momentum.
Examples:
What has greater momentum a 5 kg bowling ball moving at 3 m/s or a 0.3kg
baseball moving at 20m/s?
How fast would a 210lb running back have to move to have the same momentum as
a 280lb lineman moving at 5.0m/s?
Impulse Problems
In these types of problems focus on an object experiencing an unbalanced outside
force. The result of that unbalanced force is a change in the object’s velocity (an
acceleration) according to Newton’s second law. We limit our calculations to one
dimension allowing us to focus on the magnitude of the change in velocity. The
equation we use is a variation on Newton’s second law (F=ma) only we break a onto
Δ v/t, rearrange and get.
Ft = mΔv
 F – the unbalanced force (Newton)
 t – time (seconds)
 m – mass (kilograms)
 Δv – change in velocity (m/s)
The quantity Ft is referred to as an impulse. An impulse causes a change in
momentum (mΔv) or sometimes simply Δp. It has the same units as momentum
(kgm/s).
A force remember, is a push or a pull. During an interaction (collision) we can
increase the impulse and achieve a greater change in momentum by increasing the
time of contact, like a well struck baseball. We can also reduce the force an object
experiences by increasing the time to achieve the same change in momentum. Air
bags and seat belts in cars accomplish this during a crash.
Examples:
1. A 0.40kg softball is hit with a 180N force for 1.2 seconds. What will be the
change in the balls velocity?
2. A 0.170kg stationary hockey puck is struck by a hockey stick. After leaving
the stick the puck is moving at 20.0m/s.
a. What is the change in momentum (impulse that the puck experiences?
b. If the stick was in contact with the puck for 0.50s what was the
magnitude of the force of the stick on the puck?
c. What was the magnitude of the force of the puck on the stick?
3. A tennis ball hits a wall at 30.0m/s and rebounds at a speed of 20.0m/s if it is
in contact with the wall for 0.20s.
a. What was the momentum of the ball before hitting the wall?
b. What was the momentum of the ball after hitting the wall? (Hint:
Remember that momentum is a vector quantity.)
c. What was the change in momentum (impulse) on the ball?
d. Use the time provided to determine the magnitude of the force
exerted on the wall by the tennis ball?
e. What is the magnitude of the force exerted by the wall on the tennis
ball?
f. Explain why the wall does not experience the same change in speed
that the ball does during this event.
Three types of conserved systems.
Pbefore = Pafter
I.
Explosions:
 A system is initially at rest with no momentum.
 It is an isolated system i.e. no outside forces act on it.
 It has some sort of internal source of energy that causes movement. (the
explosion a coiled spring or chemical potential energy). Pieces of the system
will have momentum in different directions.
 When the momentum values are treated properly and added as vectors the
sum of the momentums after the explosion will still be zero. This is
conservation of momentum.
 The system is converting potential energy into kinetic energy.
Examples:
1. Basketball toss. (complete the handout for the basketball toss
demonstration.)
2. A 2.00 kg gun fires a 50.0g bullet. The bullet leaves the barrel at a speed of
300m/s. How fast will the gun recoil?
3. An astronaut has become detached from his tether and is unable to return to
his spaceship. He has a mass of 200kg. He throws a 1 kg wrench at a speed
of 12m/s in the direction away from the space ship. With what speed will he
return to the ship?
II.
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Perfectly inelastic collisions:
In these collisions the objects hit and stick together. Think of railroad cars
coupling in a stockyard.
The system is again isolated i.e. no outside forces acting on it. (We ignore
friction to make it simpler)
You need to determine the sum of all the object’s momentum before the
collision. Remember that momentum is a vector quantity so the signs will be
different if they are heading toward each other.
The value you get before the collision will be equal to the momentum of the two
objects stuck together after the collision.
Don’t forget the vector nature of momentum.
In these collisions momentum is conserved but energy is lost.
m1v1 +m2v2 = (m1+m2) v1,2
Example:
1. A 10,000kg railroad car moving at 2.0m/s collides with a stationary 8,000kg
car. The 2 cars stick together and move off. What will be their speed after
the collision?
2. An110kg linebacker is running at 3.0m/s and collides head on with a 90kg
tailback that is moving at 5.0m/s. If they stick together after the collision
what will be the speed and direction of the two-player system after the
collision?
III.
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Perfectly elastic collisions:
In these collisions the objects hit and bounce. Think pool balls.
The system is again isolated i.e. no outside forces acting on it.
You need to determine the sum of all the object’s momentum before the
collision.
It will be equal to the momentum of the larger objects momentum after the
collision.
Don’t forget the vector nature of momentum.
In these collisions both energy and momentum are conserved.
Example:
A 250g-toy car moving at 5.0m/s collides with a 200g stationary toy car. After the
collision the stationary toy car continues in the same direction at 1m/s. How fast is
the lighter car moving after the collision?